hw5soln_f09

hw5soln_f09 - ME 132 Fall 2009 Solutions to Homework 5 1 ...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ME 132 Fall 2009 Solutions to Homework 5 1. ( § 6.6, Problem 2) (a) Including the measurement noise, the closed-loop system equation becomes ˙ v ( t ) =- α m v ( t ) + E m u ( t ) + G m w ( t ) =- α m v ( t ) + E m [ K ff v des ( t )- K fb ( v ( t ) + n ( t ))] + G m w ( t ) =- α + EK fb m v ( t ) + EK ff m v des ( t ) + G m w ( t )- EK fb m n ( t ) . The two outputs are y 1 ( t ) = v ( t ) , y 2 ( t ) = u ( t ) =- K fb v ( t ) + K ff v des ( t )- K fb n ( t ) . The coefficients are then A =- α + EK fb m , B 1 = EK ff m , B 2 = G m , B 3 =- EK fb m , C 1 = 1 , D 11 = 0 , D 12 = 0 , D 13 = 0 , C 2 =- K fb , D 21 = K ff , D 22 = 0 , D 23 =- K fb . (b) The closed-loop system is stable if A < ⇔ - α + EK fb m < ⇔ α + EK fb > ⇔ K fb >- α E . (c) To determine the steady-state gain from v des to v , we set v des to a constant value v des ( t ) ≡ ¯ v des , and set the other inputs to zero, w ( t ) ≡ 0, n ( t ) ≡ 0. Then the closed-loop system equation becomes ˙ v ( t ) = Av ( t ) + B 1 ¯ v des , which has the solution v ( t ) = e At v (0) + integraldisplay t e A ( t − τ ) B 1 ¯ v des d τ = e At v (0)- B 1 A ( 1- e At ) ¯ v des . Assuming the closed-loop system is stable, the steady-state value of v ( t ) is ¯ v =- B 1 A ¯ v des . 1 Therefore, the steady-state gain from v des to v is G v des → v (0) =- B 1 A = EK ff α + EK fb . There is also a “shortcut” method of determining the steady-state gain. If the closed-loop system is stable, then we know that v ( t ) does indeed reach some steady- state value ¯ v , given constant inputs. Then we can take the limit of the closed-loop system equation as t → ∞ , 0 = A ¯ v + B 1 ¯ v des , since lim t →∞ ˙ v ( t ) = ˙ ¯ v = 0. We then obtain the same result as above, namely, ¯ v =- B 1 A ¯ v des . It is important to ensure that the closed-loop system is stable before using this shortcut method. In another problem ( § 4.5, Problem 4), we can see an example where using the shortcut method on an unstable system can lead to incorrect conclusions. (d) To determine the steady-state gain from w to v , we set w to a constant value w ( t ) ≡ ¯ w , and set the other inputs to zero, v des ( t ) ≡ 0, n ( t ) ≡ 0. Then the closed-loop system equation becomes ˙ v ( t ) = Av ( t ) + B 2 ¯ w. The steady-state value of v ( t ) is ¯ v =- B 2 A ¯ w. Therefore, the steady-state gain from w to v is G w → v (0) =- B 2 A = G α + EK fb . (e) Finally, to determine the steady-state gain from n to v , we set n to a constant value n ( t ) ≡ ¯ n , and set the other inputs to zero, v des ( t ) ≡ 0, w ( t ) ≡ 0. Then the closed-loop system equation becomes ˙ v ( t ) = Av ( t ) + B 3 ¯ n....
View Full Document

{[ snackBarMessage ]}

Page1 / 11

hw5soln_f09 - ME 132 Fall 2009 Solutions to Homework 5 1 ...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online